Normal and shear components of the stress vector

In summary, to find the normal and shear components of the stress vector on the surface ABCD, you first multiply the stress matrix by the unit normal vector to get the traction components. Then, you can use the hint given to find the normal component and the tangential component of the stress vector.
  • #1
roldy
237
2

Homework Statement


Find the normal and shear components ([itex]\vec{\sigma}_{n},\vec{\tau}[/itex]) of the stress vector on the surface ABCD.

Given:
State of stress at a point
[itex]\sigma_{xx}=3[/itex]
[itex]\sigma_{yy}=0[/itex]
[itex]\sigma_{zz}=-1[/itex]
[itex]\sigma_{xy}=-\sqrt{2}[/itex]
[itex]\sigma_{xz}=1000[/itex]

Unit normal:
[itex]\widehat{n}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}+0\widehat{k}[/itex]


Homework Equations


[itex]\vec{\sigma}_{n}=\vec{t}\cdot\widehat{n}[/itex]
[itex]\vec{t}=\vec{\sigma}_{n}+\vec{\tau}[/itex]



The Attempt at a Solution



First I take the stress matrix and multiply it by the unit normal to find the traction components

[itex]\sigma = \left[\begin{matrix}
3 & -\sqrt{2} & 1000 \\
-\sqrt{2} & 0 & -\sqrt{2} \\
1000 & -\sqrt{2} & -1 \\
\end{matrix}
\right][/itex]

[itex]\left[\begin{matrix}
3 & -\sqrt{2} & 1000 \\
-\sqrt{2} & 0 & -\sqrt{2} \\
1000 & -\sqrt{2} & -1 \\
\end{matrix}
\right]
\left[\begin{matrix}
\frac{1}{-\sqrt{2}} \\
\frac{1}{-\sqrt{2}} \\
0 \\
\end{matrix}
\right]
= \left[\begin{matrix}
t_{x} \\
t_{y} \\
t_{z} \\
\end{matrix}
\right][/itex]

This results in
[itex]\vec{t}=
\left[\begin{matrix}
\frac{3}{\sqrt{2}}-1 \\
-1 \\
\frac{1000}{\sqrt{2}}-1 \\
\end{matrix}
\right][/itex]

Then I find the normal component

[itex]\vec{\sigma}_{n}=\left[ \frac{3}{\sqrt{2}}-1,-1,\frac{1000}{\sqrt{2}}-1\right] \cdot
\left[\begin{matrix}
\frac{1}{-\sqrt{2}} \\
\frac{1}{-\sqrt{2}} \\
0 \\
\end{matrix}
\right][/itex]

[itex]\vec{\sigma}_{n} = \frac{3}{2}-\sqrt{2}[/itex]

Using the hint that was provided, I solve for [itex]\vec{\tau}[/itex]
[itex]\vec{\tau}=\vec{t}-\vec{\sigma}_n[/itex]

Is this procedure correct thus far? I've seen elsewhere that
[itex]\tau=\vec{t}-(\vec{t}\cdot \vec{n})\vec{n}[/itex]

I'm a bit confused with the notation of the supplied equation for [itex]\sigma[/itex] because if you take the dot product of a vector with a unit normal you should only get a scalar. The notation shows that it is a vector.
 
Physics news on Phys.org
  • #2
In your notation, [itex]\vec{\sigma}[/itex] is the stress tensor. The stress vector on a surface perpendicular to the unit normal [itex]\vec{n}[/itex] is given by:
[tex]\vec{t}=\vec{\sigma}\centerdot\vec{n}[/tex]
Note that [itex]\vec{\sigma}[/itex] is a second order tensor, and, when you dot it with the unit normal, you produce a vector (first order tensor) [itex]\vec{t}[/itex]. The vector [itex]\vec{t}[/itex] is not pointing perpendicular to the surface. If you want the component perpendicular to the surface, you need to dot it with the unit normal.
Scalar component of stress vector perpendicular to the surface = [itex]\vec{n}\centerdot\vec{t}
[/itex]
The normal vector component of the stress vector on the surface is equal to the scalar component times the unit normal [itex]\vec{\sigma_n}=(\vec{n}\centerdot\vec{t})\vec{n}[/itex]
The tangential vector component of the stress vector on the surface is equal to the stress vector minus the normal vector component:[itex]\vec{\tau}=\vec{t}-(\vec{n}\centerdot\vec{t})\vec{n}[/itex]

Hope this helps.

Chet
 
  • Like
Likes 1 person

1. What is the difference between normal and shear stress?

Normal stress is a force that acts perpendicular to the surface of an object, while shear stress is a force that acts parallel to the surface. In other words, normal stress pulls or pushes on an object, while shear stress causes it to slide or deform.

2. How are the normal and shear components of the stress vector calculated?

The normal and shear components of the stress vector can be calculated using the stress tensor, which is a mathematical representation of the stress at a point in a material. The normal component is determined by the normal stress acting on a given surface, while the shear component is determined by the shear stress acting on the same surface.

3. What are some real-world examples of normal and shear stress?

Normal stress can be seen in situations such as a book resting on a table, where the weight of the book creates a normal force on the table. Shear stress can be observed in objects like scissors, where the two blades slide against each other, producing a shear force.

4. How does normal and shear stress affect the strength of a material?

Normal stress can cause a material to compress or stretch, while shear stress can cause it to bend or twist. Depending on the material's properties, these stresses can either strengthen or weaken it. For example, metals are generally strong in resisting shear stress, but not as strong when it comes to normal stress.

5. Can normal and shear components of the stress vector be present at the same time?

Yes, it is possible for both normal and shear stress to be present in a material at the same time. This is often the case in real-world scenarios, where multiple forces are acting on an object. The total stress at a point is the combination of both the normal and shear components.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
937
  • Engineering and Comp Sci Homework Help
Replies
2
Views
844
  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
4
Views
286
  • MATLAB, Maple, Mathematica, LaTeX
Replies
3
Views
456
  • Advanced Physics Homework Help
Replies
6
Views
852
  • Advanced Physics Homework Help
Replies
3
Views
252
  • Advanced Physics Homework Help
Replies
8
Views
999
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top